NCERT Solutions Class 8 Maths Chapter 11 Direct and Inverse Proportions
NCERT Solutions Class 8 Maths Chapter 11 Direct and Inverse Proportions explain how two quantities change when one of them changes. Sometimes both quantities grow together. If you buy more pencils, you pay more money. This is called direct proportion. Sometimes one quantity grows and the other falls. If more workers join a job, the job finishes in fewer days. This is called inverse proportion. This chapter teaches you how to tell the two cases apart and how to find the missing value in each case.
These solutions cover every question of Exercise 11.1 and Exercise 11.2 of the NCERT (National Council of Educational Research and Training) textbook. Each answer is solved step by step in simple language, so students can follow the method and not just the final number. Teachers can use the same steps in class, and parents can use them to check homework at home. A free printable PDF of all the solutions is also available, so you can study even without the internet.
NCERT Solutions Class 8 Maths Chapter 11 Direct and Inverse Proportions
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| Number of spokes | 4 | 6 | 8 | 10 | 12 |
| Angle betweem a pair of consecutive spokes | 90º | 60º | … | … | … |
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4 hours ₹ 60
8 hours ₹ 100
12 hours ₹ 140
24 hours ₹ 180
Check if the parking charges are in direct proportion to the parking time.
NCERT Solutions for Class 8 Maths Chapter 11: All Exercises
Chapter 11 has two exercises. Exercise 11.1 is on direct proportion and Exercise 11.2 is on inverse proportion. Click on an exercise to open its detailed solutions.
| Exercise | Topic Covered | Number of Questions |
|---|---|---|
| NCERT Solutions Class 8 Maths Chapter 11: Exercise 11.1 | Direct Proportion | 10 |
| NCERT Solutions Class 8 Maths Chapter 11: Exercise 11.2 | Inverse Proportion | 11 |
NCERT Solutions Class 8 Maths Chapter 11: Exercise 11.1
Q1. Following are the car parking charges near a railway station upto:
4 hours – ₹60
8 hours – ₹100
12 hours – ₹140
24 hours – ₹180
Check if the parking charges are in direct proportion to the parking time.
Answer:
For direct proportion, the ratio of charges to hours must be the same in every case.
| Number of hours | 4 | 8 | 12 | 24 |
|---|---|---|---|---|
| Parking charges (in ₹) | 60 | 100 | 140 | 180 |
60/4 = 15
100/8 = 25/2 = 12.5
140/12 = 35/3 = 11.67 (about)
180/24 = 15/2 = 7.5
All four ratios are different from each other.
So the parking charges are not in direct proportion to the parking time.
Q2. A mixture of paint is prepared by mixing 1 part of red pigment with 8 parts of base. In the following table, find the parts of base that need to be added.
| Parts of red pigment | 1 | 4 | 7 | 12 | 20 |
|---|---|---|---|---|---|
| Parts of base | 8 | ... | ... | ... | ... |
Answer:
The ratio of red pigment to base stays the same every time, so the two are in direct proportion.
Let the unknown values be x₁, x₂, x₃ and x₄.
1/8 = 4/x₁, so x₁ = 8 × 4 = 32
1/8 = 7/x₂, so x₂ = 8 × 7 = 56
1/8 = 12/x₃, so x₃ = 8 × 12 = 96
1/8 = 20/x₄, so x₄ = 8 × 20 = 160
The completed table is:
| Parts of red pigment | 1 | 4 | 7 | 12 | 20 |
|---|---|---|---|---|---|
| Parts of base | 8 | 32 | 56 | 96 | 160 |
So the parts of base to be added are 32, 56, 96 and 160.
Q3. In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?
Answer:
Let the parts of red pigment required for 1800 mL of base be x.
| Parts of red pigment | 1 | x |
|---|---|---|
| Parts of base (in mL) | 75 | 1800 |
The parts of red pigment and the parts of base are in direct proportion.
1/75 = x/1800
75x = 1800
x = 1800/75
x = 24
Thus, 24 parts of red pigment should be mixed with 1800 mL of base.
Q4. A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Answer:
Let the number of bottles filled in five hours be x.
| Number of bottles | 840 | x |
|---|---|---|
| Time taken (in hours) | 6 | 5 |
The number of bottles and the time taken are in direct proportion.
840/6 = x/5
6x = 840 × 5
6x = 4200
x = 700
Thus, 700 bottles will be filled in 5 hours.
Q5. A photograph of a bacterium enlarged 50,000 times attains a length of 5 cm. What is the actual length of the bacterium? If the photograph is enlarged 20,000 times only, what would be its enlarged length?
Answer:
Let the actual length of the bacterium be x cm.
| Length of the bacterium (in cm) | 5 | x |
|---|---|---|
| Number of times enlarged | 50000 | 1 |
The number of times the photograph was enlarged and the length are in direct proportion.
5/50000 = x/1
x = 5/50000
x = 1/10000
x = 10⁻⁴ cm
Now let the enlarged length be y cm when the photograph is enlarged 20,000 times.
| Length of the bacterium (in cm) | 5 | y |
|---|---|---|
| Number of times enlarged | 50000 | 20000 |
5/50000 = y/20000
y = 5 × 20000 / 50000
y = 100000/50000
y = 2
The actual length of the bacterium is 10⁻⁴ cm, and its enlarged length is 2 cm.
Q6. In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?
Answer:
Let the length of the model ship be x cm.
| Height of mast | Length of ship | |
|---|---|---|
| Model ship | 9 cm | x cm |
| Actual ship | 12 m | 28 m |
The dimensions of the actual ship and the model ship are directly proportional to each other.
9/12 = x/28
12x = 9 × 28
12x = 252
x = 21
Thus, the length of the model ship is 21 cm.
Q7. Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?
Answer:
Let the distance represented on the map be x cm.
| Distance on map (in cm) | 1 | x |
|---|---|---|
| Distance on road (in km) | 18 | 72 |
The distance covered on the road and the distance shown on the map are directly proportional.
1/18 = x/72
18x = 72
x = 4
Hence, the distance represented on the map is 4 cm.
Q8. A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time
(i) the length of the shadow cast by another pole 10 m 50 cm high
(ii) the height of a pole which casts a shadow 5 m long
Answer:
First change all lengths into metres.
5 m 60 cm = 5.60 m, 3 m 20 cm = 3.20 m, 10 m 50 cm = 10.50 m
At the same time of day, the height of an object and the length of its shadow are directly proportional.
(i) Let the length of the shadow of the other pole be x m.
| Height of pole (in m) | 5.60 | 10.50 |
|---|---|---|
| Length of shadow (in m) | 3.20 | x |
5.60/3.20 = 10.50/x
5.60x = 10.50 × 3.20
5.60x = 33.6
x = 6
Hence, the length of the shadow will be 6 m.
(ii) Let the height of the pole be y m.
| Height of pole (in m) | 5.60 | y |
|---|---|---|
| Length of shadow (in m) | 3.20 | 5 |
5.60/3.20 = y/5
3.20y = 5.60 × 5
3.20y = 28
y = 8.75
Thus, the length of the shadow is 6 m and the height of the pole is 8.75 m, that is 8 m 75 cm.
Q9. A truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Answer:
Time taken by the truck to cover 14 km = 25 minutes
25 minutes in hours = 25/60 = 5/12 hour
Speed of the truck = 14 ÷ 5/12 = 14 × 12/5 = 33.6 km per hour
Distance covered in 5 hours = 33.6 × 5 = 168 km
The same answer by direct proportion:
5 hours = 5 × 60 = 300 minutes
14/25 = x/300
25x = 14 × 300 = 4200
x = 168
Therefore, the truck can travel 168 km in 5 hours at the same speed.
NCERT Solutions Class 8 Maths Chapter 11: Exercise 11.2
Q1. Which of the following are in inverse proportion?
(i) The number of workers on a job and the time to complete the job.
(ii) The time taken for a journey and the distance travelled at a uniform speed.
(iii) Area of cultivated land and the crop harvested.
(iv) The time taken for a fixed journey and the speed of the vehicle.
(v) The population of a country and the area of land per person.
Answer:
(i) It is in inverse proportion, because if there are more workers, the job takes less time.
(ii) No, it is not in inverse proportion, because in more time we cover more distance at a uniform speed.
(iii) No, it is not in inverse proportion, because more area gives more crop.
(iv) It is in inverse proportion, because with more speed we cover a fixed distance in less time.
(v) It is in inverse proportion, because if the population increases, the area of land per person decreases.
So (i), (iv) and (v) are in inverse proportion.
Q2. In a television game show, the prize money of ₹1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners.
| Number of winners | 1 | 2 | 4 | 5 | 8 | 10 | 20 |
|---|---|---|---|---|---|---|---|
| Prize for each winner (in ₹) | 1,00,000 | 50,000 | … | … | … | … | … |
Answer:
Let the unknown prizes be x₁, x₂, x₃, x₄ and x₅.
From the table:
1 × 1,00,000 = 2 × 50,000 = 1,00,000
The product stays the same, so the number of winners and the prize for each winner are inversely proportional.
1 × 1,00,000 = 4 × x₁, so x₁ = 1,00,000/4 = ₹25,000
1 × 1,00,000 = 5 × x₂, so x₂ = 1,00,000/5 = ₹20,000
1 × 1,00,000 = 8 × x₃, so x₃ = 1,00,000/8 = ₹12,500
1 × 1,00,000 = 10 × x₄, so x₄ = 1,00,000/10 = ₹10,000
1 × 1,00,000 = 20 × x₅, so x₅ = 1,00,000/20 = ₹5,000
The completed table is:
| Number of winners | 1 | 2 | 4 | 5 | 8 | 10 | 20 |
|---|---|---|---|---|---|---|---|
| Prize for each winner (in ₹) | 1,00,000 | 50,000 | 25,000 | 20,000 | 12,500 | 10,000 | 5,000 |
The prize money given to an individual winner is inversely proportional to the number of winners.
Q3. Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table.
| Number of spokes | 4 | 6 | 8 | 10 | 12 |
|---|---|---|---|---|---|
| Angle between a pair of consecutive spokes | 90° | 60° | … | … | … |
(i) Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion?
(ii) Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.
(iii) How many spokes would be needed if the angle between a pair of consecutive spokes is 40°?
Answer:
Let the unknown angles be x₁, x₂ and x₃.
From the given table:
4 × 90° = 360° = 6 × 60°
(i) The product is the same, so the number of spokes and the angle between a pair of consecutive spokes are inversely proportional to each other.
Now we find x₁, x₂ and x₃.
4 × 90° = 8 × x₁, so x₁ = (4 × 90°)/8 = 45°
x₂ = (4 × 90°)/10 = 36°
x₃ = (4 × 90°)/12 = 30°
The completed table is:
| Number of spokes | 4 | 6 | 8 | 10 | 12 |
|---|---|---|---|---|---|
| Angle between a pair of consecutive spokes | 90° | 60° | 45° | 36° | 30° |
(ii) Let the angle on a wheel with 15 spokes be x.
4 × 90° = 15 × x
x = 360°/15 = 24°
Hence, the angle between a pair of consecutive spokes on a wheel with 15 spokes is 24°.
(iii) Let the number of spokes be y when the angle is 40°.
4 × 90° = y × 40°
y = 360/40 = 9
Hence, 9 spokes would be needed.
Q4. If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get if the number of the children is reduced by 4?
Answer:
Total number of children = 24
If the number of children is reduced by 4, the remaining children = 24 − 4 = 20
Let the number of sweets each of the 20 children will get be x.
| Number of children | 24 | 20 |
|---|---|---|
| Number of sweets each | 5 | x |
If the number of children is reduced, each child will get more sweets. So this is a case of inverse proportion.
24 × 5 = 20 × x
120 = 20x
x = 6
Hence, each child will get 6 sweets.
Q5. A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle?
Answer:
Total number of animals is 20.
If there are 10 more animals, the total number of animals would be 30.
Let the food last for x days.
| Number of animals | 20 | 30 |
|---|---|---|
| Number of days | 6 | x |
The food lasts longer if there are fewer animals, so the number of animals and the number of days are inversely proportional.
20 × 6 = 30 × x
120 = 30x
x = 4
Therefore, the food will last for 4 days.
Q6. A batch of bottles was packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?
Answer:
Let the number of boxes filled by using 20 bottles in each box be x.
| Number of bottles in each box | 12 | 20 |
|---|---|---|
| Number of boxes | 25 | x |
If more bottles are put in each box, fewer boxes are needed. So the two are inversely proportional.
12 × 25 = 20 × x
x = (12 × 25)/20 = 300/20 = 15
Hence, the number of boxes required to pack 20 bottles in each box is 15.
Q7. A factory requires 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days?
Answer:
Let the number of machines required to finish the work in 54 days be x.
| Number of machines | 42 | x |
|---|---|---|
| Number of days | 63 | 54 |
If more machines are used, fewer days are needed. So the number of machines and the number of days are inversely proportional.
42 × 63 = x × 54
2646 = 54x
x = 2646/54 = 49
Hence, 49 machines would be required to produce the same number of articles in 54 days.
Q8. A car takes 2 hours to reach a destination by travelling at the speed of 60 km/h. How long will it take when the car travels at the speed of 80 km/h?
Answer:
Let the time taken at a speed of 80 km/h be x hours.
| Speed (in km/h) | 60 | 80 |
|---|---|---|
| Time taken (in hours) | 2 | x |
If the speed is more, the car takes less time. So speed and time are inversely proportional.
60 × 2 = 80 × x
120 = 80x
x = 120/80 = 3/2
Hence, the car will take 3/2 hours, that is 1½ hours or 1 hour 30 minutes.
Q9. Two persons could fit new windows in a house in 3 days.
(i) One of the persons fell ill before the work started. How long would the job take now?
(ii) How many persons would be needed to fit the windows in one day?
Answer:
(i) Since one person fell ill before the work started, only one person is left.
Let the number of days required by 1 person be x.
| Number of persons | 2 | 1 |
|---|---|---|
| Number of days | 3 | x |
If fewer persons work, more days are needed. So this is a case of inverse proportion.
2 × 3 = 1 × x
x = 6
Hence, 1 person will take 6 days to fit all the windows.
(ii) Let the number of persons needed to fit all the windows in one day be y.
| Number of persons | 2 | y |
|---|---|---|
| Number of days | 3 | 1 |
If fewer days are given, more persons are needed. So this is again inverse proportion.
2 × 3 = y × 1
y = 6
Hence, 6 persons are required to fit all the windows in one day.
Q10. A school has 8 periods a day, each of 45 minutes duration. How long would each period be if the school has 9 periods a day, assuming the number of school hours to be the same?
Answer:
Let the duration of each period, when there are 9 periods a day, be x minutes.
| Number of periods | 8 | 9 |
|---|---|---|
| Duration of each period (in minutes) | 45 | x |
If there are more periods in the same school time, each period will be shorter. So this is a case of inverse proportion.
45 × 8 = x × 9
360 = 9x
x = 40
Hence, the duration of each period will be 40 minutes.